Answer:

To solve the problem, we can use the equations of motion for projectile motion.

Given:

Angle of projection (θ) = 35°

Horizontal distance (range) (R) = 30 m

Vertical height (h) = 1.8 m

a) To find the initial velocity of the ball (u), we can use the equations:

R = (u² * sin(2θ)) / g

h = (u² * sin²(θ)) / (2g)

where g is the acceleration due to gravity.

Substituting the given values into the equations:

30 = (u² * sin(2 * 35°)) / g

1.8 = (u² * sin²(35°)) / (2g)

We can divide these two equations to eliminate u²:

(30 / 1.8) = (sin(2 * 35°) / (2 * sin²(35°)))

Simplifying the equation further, we find:

(30 / 1.8) = 2 * sin(35°) / (1 - cos(2 * 35°))

Now we can solve for sin(35°):

sin(35°) = (30 / 1.8) * (1 - cos(2 * 35°)) / 2

Using a scientific calculator, we can find the value of sin(35°) ≈ 0.5736.

Substituting this value back into the equation, we have:

u² = (30 / 1.8) * (1 - cos(2 * 35°)) / (2 * 0.5736)

Solving this equation, we find:

u ≈ 15.55 m/s (approximately)

Therefore, the initial velocity of the ball should be approximately 15.55 m/s.

b) To find the time taken for the ball to reach the target, we can use the equation for horizontal distance (R) and the horizontal component of velocity (uₓ):

R = uₓ * t

We already know R = 30 m, so we need to find uₓ. Using the equation:

uₓ = u * cos(θ)

where u is the initial velocity and θ is the angle of projection.

Substituting the given values into the equation:

uₓ = 15.55 m/s * cos(35°)

Calculating this value, we find:

uₓ ≈ 12.71 m/s (approximately)

Now we can use the equation to find the time (t):

30 m = 12.71 m/s * t

Solving for t, we have:

t ≈ 2.36 s (approximately)

Therefore, the time taken for the ball to reach the target is approximately 2.36 seconds.

Explanation: